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Chapter 12: Problem 37

Rewrite each equation in the standard form for the equation of a circle, and identify its center and radius. $$x^{2}-2 x+y^{2}-4 y-3=0$$

Short Answer

Expert verified
The center is (1, 2) and the radius is \2 \sqrt 2\.

Step by step solution

01

Group the variables

Group the x-terms and the y-terms together. The given equation is \[ x^2 - 2x + y^2 - 4y - 3 = 0 \]
02

Move the constant to the other side

Add 3 to both sides of the equation to isolate the constant term:\[ x^2 - 2x + y^2 - 4y = 3 \]
03

Complete the square for the x-terms

To complete the square for the x-terms, take the coefficient of x (which is -2), divide it by 2 to get -1, and then square it to get 1. Add and subtract this value inside the equation:\[ x^2 - 2x + 1 - 1 + y^2 - 4y = 3 \]\[ (x - 1)^2 - 1 + y^2 - 4y = 3 \]
04

Complete the square for the y-terms

To complete the square for the y-terms, take the coefficient of y (which is -4), divide it by 2 to get -2, and then square it to get 4. Add and subtract this value inside the equation:\[ (x - 1)^2 - 1 + y^2 - 4y + 4 - 4 = 3 \]\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 3 \]
05

Simplify the equation

Combine like terms:\[ (x - 1)^2 + (y - 2)^2 - 5 = 3 \]Add 5 to both sides to rewrite it in the standard form of a circle:\[ (x - 1)^2 + (y - 2)^2 = 8 \]
06

Identify the center and radius

The standard form of a circle is \[ (x - h)^2 + (y - k)^2 = r^2 \]By comparing it with our equation, we get the center \((h, k) = (1, 2)\) and \(r^2 = 8\)So, the radius \( r \) is \sqrt{8}\ or \ 2\sqrt{2}.\

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

completing the square
Completing the square is a method used to rewrite quadratic equations into a more manageable form. This is especially useful in the context of circles. To complete the square:
  • Identify the quadratic term (like x虏 or y虏) and its linear term (like -2x or -4y).
  • Take the coefficient of the linear term, divide it by 2, and then square the result.
  • Add and subtract this squared value within the equation to complete the square.
For example, consider the x-terms in the equation x虏 - 2x.
  • Take the coefficient of x, which is -2.
  • Divide it by 2 to get -1.
  • Square -1 to get 1.
So, you add and subtract 1 within the equation to form a perfect square: x虏 - 2x + 1 - 1 = (x - 1)虏 - 1.
Similarly, for the y-terms in the equation y虏 - 4y:
  • Take the coefficient of y, which is -4.
  • Divide it by 2 to get -2.
  • Square -2 to get 4.
Thus, you add and subtract 4 within the equation to form: y虏 - 4y + 4 - 4 = (y - 2)虏 - 4.
circle standard form
The standard form of a circle鈥檚 equation is \((x - h)虏 + (y - k)虏 = r虏\). Here, \(h\) and \(k\) represent the coordinates of the circle's center, and \(r\) represents the radius.
Let's restructure the given equation \(x虏 - 2x + y虏 - 4y - 3 = 0\) step by step in order to rewrite it in standard form:
  • Group the x-terms and y-terms: \(x虏 - 2x + y虏 - 4y = 3 \).
  • Complete the square separately for x and y terms to transform them: \((x - 1)虏 - 1 + (y - 2)虏 - 4 = 3\).
  • Combine like terms and rearrange: \((x - 1)虏 + (y - 2)虏 = 8\).
Now, compare this with the general form \((x - h)虏 + (y - k)虏 = r虏 \). It's evident that our circle has its center at \(h = 1\) and \(k = 2\), while its radius squared, \(r虏=8\).
radius and center calculation
Finding the radius and center of a circle is straightforward once the equation is in standard form. The standard form of a circle鈥檚 equation is \((x - h)虏 + (y - k)虏 = r虏\).
Here's how to identify the center and radius from the standard form:
  • The center of the circle is given by the point \( (h, k) \).
  • The radius \ (r) \ is the square root of the constant on the right side of the equation, \ (r = \sqrt{r虏}) \.
For the equation we derived \((x - 1)虏 + (y - 2)虏 = 8\):
  • We have \( (h, k) = (1, 2)\), so the center of the circle is at the point (1, 2).
  • Next, since \(r虏 = 8\), the radius is \sqrt{8} \ or \ 2\sqrt{2} \.
The center and radius are crucial as they determine the circle's position and size. Understanding these allows you to graph the circle accurately or apply further geometric concepts.

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Most popular questions from this chapter

Graph both equations of each system on the same coordinate axes. Use elimination of variables to find all points of intersection. $$ \begin{aligned} &x^{2}+\frac{y^{2}}{9}=1\\\ &x^{2}+y^{2}=4 \end{aligned} $$

$$\text {Graph each equation.}$$ $$x^{2}+y^{2}=0$$

$$\text {Graph each equation.}$$ $$y=-\sqrt{1-x^{2}}$$

Graph \(y_{1}=\sqrt{x^{2}-1}, y_{2}=-\sqrt{x^{2}-1}, y_{3}=x,\) and \(y_{4}=-x\) to get the graph of the hyperbola \(x^{2}-y^{2}=1\) along with its asymptotes. Use the viewing window \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3 .\) Notice how the branches of the hyperbola approach the asymptotes.

Determine whether the ordered pair \((3,-4)\) satisfies each system of inequalities. $$\begin{aligned}&4 x^{2}+y^{2} \leq 36\\\&x^{2}+y^{2} \geq 25\end{aligned}$$
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